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Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2009, Article ID 732106, 28 pages
doi:10.1155/2009/732106
Research Article
Inequalities for Single Crystal Tube Growth by
Edge-Defined Film-Fed Growth Technique
Stefan Balint
1
and Agneta M. Balint
2
1
Department of Computer Science, Faculty of Mathematics and Computer Science,
West University of Timisoara, Blv. V.Parvan 4, 300223 Timisoara, Romania
2
Faculty of Physics, West University of Timisoara, Blv. V.Parvan 4, 300223 Timisoara, Romania
Correspondence should be addressed to Agneta M. Balint,
Received 3 January 2009; Accepted 29 March 2009
Recommended by Yong Zhou
The axi-symmetric Young-Laplace differential equation is analyzed. Solutions of this equation can
describe the outer or inner free surface of a static meniscus the static liquid bridge free surface
between the shaper and the crystal surface occurring in single crystal tube growth. The analysis
concerns the dependence of solutions of the equation on a parameter p which represents the
controllable part of the pressure difference across the free surface. Inequalities are established for p
which are necessary or sufficient conditions for the existence of solutions which represent a stable
and convex outer or inner free surfaces of a static meniscus. The analysis is numerically illustrated
for the static menisci occurring in silicon tube growth by edge-defined film-fed growth EFGs
technique. This kind of inequalities permits the adequate choice of the process parameter p.With
this aim this study was undertaken.
Copyright q 2009 S. Balint and A. M. Balint. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
The first successful Si tube growth was reported in 1. Also a theory of tube growth by E.F.G.
process is developed there to show the dependence of the tube wall thickness on the growth
variables. The theory concerns the calculation of the shape of the liquid-vapor interface i.e.,
the free surface of the meniscus and of the heat flow in the system. The inner and the outer
free surface shapes of the meniscus Figure 1 were calculated from Young-Laplace capillary
equation, in which the pressure difference Δp across a point on the free surface is considered
to be Δp ρ · g · H
eff
constant, where H
eff
represents the effective height of the growth
interface Figure 1. The above approximation of Δp is valid when H
eff
h, where h is
the height of the growth interface above the shaper top. Another approximation used in
1 is that the outer and inner free surface shapes are approximated by circular segments.
With these relatively tight tolerances concerning the menisci in conjunction with the heat
2 Journal of Inequalities and Applications
flow calculation in the system, the predictive model developed in 1 has been shown to be
useful tool understanding the feasible limits of the wall thickness control. A more accurate
predictive model would require an increase of the acceptable tolerance range introduced by
approximation.
The growth process was scaled up by Kaljes et al. in 2 to grow 15 cm diameter silicon
tubes. It has been realized that theoretical investigations are necessary for the improvement
of the technology. Since the growth system consists of a small die type 1mm width and
athintubeorder of 200 μm wall thickness, the width of the melt/solid interface and the
meniscus are accordingly very small. Therefore, it is essential to obtain accurate solution for
the free surface of the meniscus, the temperature, and the liquid-crystal interface position in
this tinny region.
For single crystal tube growth by edge-defined film-fed growth E.F.G. technique, in
hydrostatic approximation the free surface of a static meniscus is described by the Young-
Laplace capillary equation 3:
γ ·
1
R
1
1
R
2
ρ · g · z − p. 1.1
Here γ is the melt surface tension, ρ denotes the melt density, g is the gravity acceleration,
1/R
1
, 1/R
2
denote the main normal curvatures of the free surface at a point M of the free
surface, z is the coordinate of M with respect to the Oz axis, directed vertically upwards,
and p is the pressure difference across the free surface. For the outer free surface, p p
e
p
m
− p
e
g
− ρ · g · H and for the inner free surface, p p
i
p
m
− p
i
g
− ρ · g · H.
Here p
m
denotes the hydrodynamic pressure in the meniscus melt, p
e
g
,p
i
g
denote the
pressure of the gas flow introduced in the furnace in order to release the heat from the outer
and inner walls of the tube, respectively, and H denotes the melt column height between the
horizontal crucible melt level and the shaper top level. When the shaper top level is above
the crucible melt level, then H>0, and when the crucible melt level is above the shaper top
level, then H<0 see Figure 1.
To calculate the outer and inner free surface shapes of the static meniscus, it is
convenient to employ the Young-Laplace 1.1 in its differential form. This form of the 1.1
can be obtained as a necessary condition for the minimum of the free energy of the melt
column 3.
For the growth of a single crystal tube of inner radius r
i
∈ R
gi
, R
gi
R
ge
/2 and
outer radius r
e
∈ R
gi
R
ge
/2,R
ge
the axi-symmetric differential equation of the outer free
surface is given by
z
ρ · g ·z −p
e
γ
1
z
2
3/2
−
1
r
·
1
z
2
· z
for r ∈
r
e
,R
ge
, 1.2
which is the Euler equation for the energy functional
I
e
z
R
ge
r
e
γ ·
1
z
2
1/2
1
2
· ρ · g · z
2
− p
e
· z
· r · dr,
z
r
e
h
e
> 0,z
R
ge
0.
1.3
Journal of Inequalities and Applications 3
z
Inner gas
flow
Outer gas
flow
r
i
r
e
Tube
Outer
free surface
Inner
free surface
Meniscus melt
Shaper
Capillary channel
Crucible melt
R
gi
R
ge
H
H
eff
r
h
i
h
e
0
z
i
r
z
e
r
α
c
α
c
α
g
α
g
Figure 1: Axisymmetric meniscus geometry in the tube growth by E.F.G. method.
The axi-symmetric differential equation of the inner free surface is given by
z
ρ · g ·z −p
i
γ
1
z
2
3/2
−
1
r
·
1
z
2
· z
for r ∈
R
gi
,r
i
, 1.4
which is the Euler equation for the energy functional
I
i
z
r
i
R
gi
γ ·
1
z
2
1/2
1
2
· ρ · g · z
2
− p
i
· z
· r · dr,
z
R
gi
0,z
r
i
h
i
> 0.
1.5
The state of the arts at the time 1993-1994, concerning the dependence of the shape of the
meniscus free surface on the pressure difference p for small and large bond numbers, in
the case of the growth of single crystal rods by E.F.G. technique, are summarized in 4.
According to 4, for the general differential equation 1.2, 1.4 describing the free surface
of a liquid meniscus, there are no complete analysis and solution. For the general equation
only numerical integrations were carried out for a number of process parameter values that
were of practical interest at the moment.
Later, in 2001, Rossolenko shows in 5 that the hydrodynamic factor is too small
to be considered in the automated single crystal tube growth. Finally, in 6 the authors
present theoretical and numerical study of meniscus dynamics under axi-symmetric and
asymmetric configurations. In 6 the meniscus free surface is approximated by an arc of
4 Journal of Inequalities and Applications
constant curvature, and a meniscus dynamics model is developed to consider meniscus
shape and its dynamics, heat and mass transfer around the die-top and meniscus. Analysis
reveals the correlations among tube wall thickness, effective melt height, pull-rate, die-top
temperature, and crystal environmental temperature.
In the present paper the shape of the inner and outer free surfaces of the static meniscus
is analyzed as function of p, the controllable part of the pressure difference across the free
surface, and the static stability of the free surfaces is investigated. The novelty with respect
to the considerations presented in literature consists in the fact that the free surface is not
approximated as in 1, 6, by an arc with constant curvature, and the pressure of the gas flow
introduced in the furnace for releasing the heat from the tube wall is taken in consideration.
The setting of the thermal conditions is not considered in this paper.
2. Meniscus Outer Free Surface Shape Analysis in
the Case of Tube Growth
Consider the differential equation
z
ρ · g ·z −p
e
γ
1
z
2
3/2
−
1
r
·
1
z
2
· z
for
R
gi
R
ge
2
≤ r ≤ R
ge
, 2.1
and α
c
,α
g
such that 0 <α
c
<π/2 −α
g
,α
g
∈ 0,π/2.
Definition 2.1. A solution z zx of the 2.1 describes the outer free surface of a static
meniscus on the interval r
e
,R
ge
R
gi
R
ge
/2 <r
e
<R
ge
if possesses the following
properties:
a z
r
e
−tanπ/2 −α
g
,
b z
R
ge
−tan α
c
, and
c zR
ge
0andzr is strictly decreasing on r
e
,R
ge
.
The described outer free surface is convex on r
e
,R
ge
if in addition the following inequality
holds:
d z
r > 0 ∀r ∈ r
e
,R
ge
.
Theorem 2.2. If there exists a solution of 2.1, which describes a convex outer free surface of a static
meniscus on the closed interval r
e
,R
ge
, then the following inequalities hold:
− γ ·
π/2 −
α
c
α
g
R
ge
− r
e
· cos α
c
γ
R
ge
· sin α
c
≤ p
e
≤−γ ·
π/2 −
α
c
α
g
R
ge
− r
e
· sin α
g
ρ · g ·
R
ge
− r
e
· tan
π
2
− α
g
γ
r
e
· cos α
g
.
2.2
Journal of Inequalities and Applications 5
Proof. Let zr be a solution of 2.1, which describes a convex outer free surface of a static
meniscus on the closed interval r
e
,R
ge
and αr−arctanz
r. The function αr verifies
the equation
α
r
p
e
− ρ · g · z
r
γ
·
1
cos α
r
−
1
r
· tan α
r
2.3
and the boundary conditions
α
r
e
π
2
− α
g
,α
R
ge
α
c
. 2.4
Hence, there exists r
∈ r
e
,R
ge
such that the following equality holds:
p
e
−γ ·
π/2 −
α
c
α
g
R
ge
− r
e
· cos α
r
ρ · g · z
r
γ
r
· sin α
r
. 2.5
Since z
r > 0onr
e
,R
ge
, z
r is strictly increasing and αr−arctan z
r is strictly
decreasing on r
e
,R
ge
, therefore the values of αr are in the range α
c
,π/2 − α
g
and the
following inequalities hold:
α
c
≤ α
r
≤
π
2
− α
g
,
sin α
g
≤ cos α
r
≤ cos α
c
,
sin α
c
≤ sin α
r
≤ cos α
g
,
−tan
π
2
− α
g
≤ z
r
≤−tanα
c
,
ρ · g ·
R
ge
− r
· tan α
c
≤ ρ · g · z
r
≤ ρ · g ·
R
ge
− r
· tan
π
2
− α
g
.
2.6
Equality 2.5 and inequalities 2.6 imply inequalities 2.2.
Corollary 2.3. If r
e
R
ge
/n with 1 <n<2 · R
ge
/R
gi
R
ge
, then inequalities 2.2 become
− γ ·
π/2 −
α
c
α
g
R
ge
·
n
n − 1
· cos α
c
γ
R
ge
· sin α
c
≤ p
e
≤−γ ·
π/2 −
α
c
α
g
R
ge
·
n
n − 1
· sin α
g
ρ · g ·R
ge
·
n − 1
n
· tan
π
2
− α
g
γ
R
ge
· n · cos α
g
.
2.7
6 Journal of Inequalities and Applications
Corollary 2.4. If n → 2 · R
ge
/R
gi
R
ge
,thenr
e
→ R
gi
R
ge
/2 and 2.7 becomes
− 2 ·γ ·
π/2 −
α
c
α
g
R
ge
− R
gi
· cos α
c
γ
R
ge
· sin α
c
≤ p
e
≤−2 ·γ ·
π/2 −
α
c
α
g
R
ge
− R
gi
· sin α
g
ρ · g ·
R
ge
− R
gi
2
· tan
π
2
− α
g
2 · γ
R
gi
R
ge
· cos α
g
.
2.8
If n → 1,thenr
e
→ R
ge
and p
e
→−∞.
Theorem 2.5. Let n be such that 1 <n<2 ·R
ge
/R
gi
R
ge
.Ifp
e
satisfies the inequality
p
e
< −γ ·
π/2 −
α
c
α
g
R
ge
·
n
n − 1
· cos α
c
γ
R
ge
· sin α
c
, 2.9
then there exists r
e
in the closed interval R
ge
/n, R
ge
such that the solution of the initial value
problem
z
ρ · g ·z −p
e
γ
·
1
z
2
3/2
−
1
r
·
1 z
2
· z
for
R
gi
R
ge
2
<r≤ R
ge
z
R
ge
0,z
R
ge
−tan α
c
, 0 <α
c
<
π
2
− α
g
,α
g
∈
0,
π
2
2.10
on the interval r
e
,R
ge
describes the convex outer free surface of a static meniscus.
Proof. Consider the solution zr of the initial value problem 2.10. Denote by I the maximal
interval on which the function zr exists and by αr the function αr−arctan z
r
defined on I. Remark that for αr the equality 2.3 holds.
Since
z
R
ge
> 0,z
R
ge
−tan α
c
< 0,z
R
ge
> −tan
π
2
− α
g
, 2.11
there exists r
∈ I,0<r
<R
ge
such that for any r ∈ r
,R
ge
the following inequalities hold:
z
r
> 0,z
r
≤−tanα
c
,z
r
≥−tan
π
2
− α
g
. 2.12
Let r
∗
be defined by
r
∗
inf
r
∈ I | 0 <r
<R
ge
such that for any r ∈
r
,R
ge
inequalities
2.12
hold
. 2.13
It is clear that r
∗
≥ 0 and for any r ∈ r
∗
,R
ge
inequalities 2.12 hold.
Journal of Inequalities and Applications 7
From 2.12 and 2.13 it follows that z
r is strictly increasing and bounded on
r
∗
,R
ge
. Therefore z
r
∗
0lim
r →r
∗
,r>r
∗
z
r exists and satisfies
−tan
π
2
− α
g
≤ z
r
∗
0
≤−tanα
c
. 2.14
Moreover, since zr is strictly decreasing and possesses bounded derivative on r
∗
,R
ge
,
zr
∗
0lim
r →r
∗
,r>r
∗
zr exists too, it is finite, and satisfies
0 <
R
ge
− r
∗
· tan α
c
≤ z
r
∗
0
≤
R
ge
− r
∗
· tan
π
2
− α
g
< ∞. 2.15
We will show now that r
∗
>R
ge
/n and z
r
∗
0−tanπ/2 − α
g
. In order to show that
r
∗
>R
ge
/n we assume the contrary, that is, that r
∗
≤ R
ge
/n . Under this hypothesis we
have
α
r
∗
0
− α
R
ge
−α
r
·
R
ge
− r
∗
−
p
e
γ
ρ · g ·z
r
γ
sin α
r
r
·
R
ge
− r
∗
cos α
r
>
π/2 −
α
c
α
g
R
ge
·
n
n − 1
· cos α
c
−
1
R
ge
· sin α
c
ρ · g ·z
r
γ
sin α
r
r
·
R
ge
− r
∗
cos α
r
>
π
2
−
α
c
α
g
2.16
for some r
∈ r
∗
,R
ge
. Hence αr
∗
0 >π/2 − α
g
. This last inequality is impossible, since
according to the inequality 2.14, we have αr
∗
0 ≤ π/2 − α
g
. Therefore, r
∗
, defined by
2.14,satisfiesr
∗
>R
ge
/n.
In order to show that z
r
∗
0−tanπ/2 − α
g
we remark that from the definition
2.14 of r
∗
it follows that at least one of the following three equalities holds:
z
r
∗
0
−tan α
c
,z
r
∗
0
−tan
π
2
− α
g
,z
r
∗
0
0. 2.17
Since z
r
∗
0 <z
r ≤−tan α
c
for any r ∈ r
∗
,R
ge
it follows that the equality z
r
∗
0
−tan α
c
is impossible. Hence, we obtain that at r
∗
at least one of the following two equalities
holds: z
r
∗
00, z
r
∗
0−tanπ/2 −α
g
.Weshownowthattheequalityz
r
∗
00
is impossible. For that we assume the contrary, that is, z
r
∗
00. Under this hypothesis,
8 Journal of Inequalities and Applications
from 2.12 we have:
p
e
g · ρ · z
r
∗
0
γ
r
∗
· sin α
r
∗
0
>g· ρ ·
R
ge
− r
∗
· tan α
c
γ
R
ge
· sin α
c
>
γ
R
ge
· sin α
c
− γ ·
π/2 −
α
c
α
g
R
ge
·
n
n − 1
· cos α
c
>p
e
2.18
what is impossible.
In this way we obtain that the equality z
r
∗
0−tanπ/2 − α
g
holds.
For r
e
r
∗
the solution of the initial value problem 2.8 on the interval r
e
,R
ge
describes a convex outer free surface of a static meniscus.
Corollary 2.6. If for p
e
the following inequality holds:
p
e
< −2 · γ ·
π/2 −
α
c
α
g
R
ge
− R
gi
· cos α
c
γ
R
ge
· sin α
c
, 2.19
then there exists r
e
in the interval R
gi
R
ge
/2,R
ge
such that the solution of the initial value
problem 2.10 on the interval r
e
,R
ge
describes a convex outer free surface of a static meniscus.
Corollary 2.7. If for 1 <n
<n<2 · R
ge
/R
gi
R
ge
the following inequalities hold:
− γ ·
π/2 −
α
c
α
g
R
ge
·
n
n
− 1
· sin α
g
ρ · g · R
ge
·
n
− 1
n
· tan
π
2
− α
g
γ
R
ge
· n
cos α
g
<p
e
< −γ ·
π/2 −
α
c
α
g
R
ge
·
n
n − 1
· cos α
c
γ
R
ge
· sin α
c
,
2.20
then there exists r
e
in the interval R
ge
/n, R
ge
/n
such that the solution of the initial value problem
2.10 on the interval r
e
,R
ge
describes a convex outer free surface of a static meniscus. The existence
of r
e
and the inequality r
e
≥ R
ge
/n follows from Theorem 2.5. The inequality r
e
≤ R
ge
/n
follows
from Corollary 2.3.
Remark 2.8. The solution of the initial value problem 2.10 is convex at R
ge
i.e., z
R
ge
> 0
if and only if
p
e
<
γ
R
ge
· sin α
c
. 2.21
Journal of Inequalities and Applications 9
That is because z
R
ge
> 0 if and only if α
R
ge
−z
R
ge
· cos
2
α
c
< 0, that is,
p
e
γ
−
sin α
c
R
ge
< 0 ⇐⇒ p
e
<
γ
R
ge
· sin α
c
. 2.22
Moreover, if p
e
<γ/R
ge
· sin α
c
,thesolutionzr of the initial value problem 2.8 is convex
everywhere i.e., z
r > 0forr ∈ I, 0 <r≤ R
ge
. That is because the change of convexity
implies the existence of r
∈ I, 0 <r
<R
ge
such that αr
>α
c
,zr
> 0andp
e
ρ ·g ·zr
γ/r
· sin αr
>γ/R
ge
· sin α
c
, what is impossible.
Theorem 2.9. If a solution z
1
z
1
r of 2.1 describes a convex outer free surface of a static
meniscus on the interval r
e
,R
ge
R
gi
R
ge
/2 <r
e
<R
ge
, then it is a weak minimum for the
energy functional of the melt column:
I
e
z
R
ge
r
e
γ ·
1
z
2
1/2
1
2
· ρ · g · z
2
− p
e
· z
· r · dr
z
r
e
z
1
r
e
,z
R
ge
z
1
R
ge
0.
2.23
Proof. Since 2.1 is the Euler equation for 2.23,itissufficient to prove that the Legendre
and Jacobi conditions are satisfied in this case.
Denote by Fr, z, z
the function defined as
F
r, z, z
r ·
1
2
· ρ · g · z
2
− p
e
· z γ ·
1
z
2
1/2
. 2.24
It is easy to verify that we have
∂
2
F
∂z
2
r · γ
1
z
2
3/2
> 0. 2.25
Hence, the Legendre condition is satisfied.
The Jacobi equation
∂
2
F
∂z
2
−
d
dr
∂
2
F
∂z ∂z
· η −
d
dr
∂
2
F
∂z
2
· η
0 2.26
in this case is given by
d
dr
⎛
⎜
⎝
r · γ
1
z
2
3/2
· η
⎞
⎟
⎠
− ρ · g · r ·η 0. 2.27
10 Journal of Inequalities and Applications
For 2.27 the following inequalities hold:
r · γ
1
z
2
3/2
≥ r · γ · cos
3
π
2
− α
g
r · γ · sin
3
α
g
, −ρ · g · r ≤ 0. 2.28
Hence, the equation
η
· r · γ · sin
3
α
g
0 2.29
is a Sturm type upper bound for 2.277.
Since every nonzero solution of 2.29 vanishes at most once on the interval r
e
,R
ge
,
the solution ηr of the initial value problem
d
dr
⎛
⎜
⎝
r · γ
1
z
2
3/2
· η
⎞
⎟
⎠
− ρ · g · r ·η 0,
η
r
e
0,η
r
e
1
2.30
has only one zero on the interval r
e
,R
ge
7. Hence the Jacobi condition is satisfied.
Definition 2.10. A solution z zr of 2.1 which describes the outer free surface of a static
meniscus is said to be stable if it is a weak minimum of the energy functional of the melt
column.
Remark 2.11. Theorem 2.9 shows that if z zr describes a convex outer free surface of a
static meniscus on the interval r
e
,R
ge
, then it is stable.
Theorem 2.12. If the solution z zr of the initial value problem 2.10 is concave (i.e., z”r < 0)
on the interval r
e
,R
ge
R
gi
R
ge
/2 <r
e
<R
ge
, then it does not describe the outer free surface of
a static meniscus on r
e
,R
ge
.
Proof. z
r < 0onr
e
,R
ge
implies that z
r is strictly decreasing on r
e
,R
ge
. Hence z
r
e
>
z
R
ge
−tan α
c
> −tanπ/2 −α
g
.
Theorem 2.13. If p
e
>γ/R
ge
·sin α
c
and there exists r
e
∈ R
gi
R
ge
/2,R
ge
such that the solution
of the initial value problem 2.10 is the outer free surface of a static meniscus on r
e
,R
ge
, then for p
e
the following inequalities hold:
γ
R
ge
· sin α
c
<p
e
≤ ρ · g ·
R
ge
− r
e
· tan
π
2
− α
g
γ
r
e
· cos α
g
. 2.31
Proof. Denote by zr the solution of the initial value problem 2.10 which is assumed to
represent the outer free surface of a static meniscus on the closed interval r
e
,R
ge
.Letαr
be defined as in Theorem 2.2.forr ∈ r
e
,R
ge
. Since p
e
>γ/R
ge
· sin α
c
, we have z
R
ge
−1/cos
2
α
c
· α
R
ge
< 0. Hence α
R
ge
> 0 and therefore αr <αR
ge
α
c
for r<R
ge
, r
Journal of Inequalities and Applications 11
close to R
ge
. Taking into account the fact that αr
e
π/2 −α
e
>α
c
it follows that there exists
r
∈ r
e
,R
ge
such that α
r
0.
Therefore p
e
ρ · g · zr
γ/r
· sin αr
. Since 0 ≤ αr
≤ π/2 − α
g
and r
e
<r
,the
following inequality holds: γ/r
· sin αr
<γ/r
e
· cos α
g
. On the other hand zr
<zr
e
≤
R
ge
− r
e
· tanπ/2 −α
g
. Using the above evaluations we obtain inequalities 2.31.
Remark 2.14. If r
e
appearing in Theorem 2.13 is represented as r
e
R
ge
/n,1 <n<
2 · R
ge
/R
gi
R
ge
, then inequality 2.31 becomes
γ
R
ge
· sin α
c
<p
e
≤ ρ · g ·
n − 1
n
· R
ge
· tan
π
2
− α
g
γ
R
ge
· n · cos α
g
. 2.32
For n → 2 ·R
ge
/R
gi
R
ge
inequality 2.32 becomes
γ
R
ge
· sin α
c
<p
e
≤ ρ · g ·
R
ge
− R
gi
2 · R
ge
· tan
π
2
− α
g
2 · γ
R
gi
R
ge
· cos α
g
2.33
Theorem 2.15. Let n be 1 <n<2 · R
ge
/R
gi
R
ge
.Ifforp
e
the following inequality holds:
p
e
>ρ·g · R
ge
·
n − 1
n
· tan α
c
n
R
ge
· γ, 2.34
then the solution zr of the initial value problem 2.10 is concave on the interval I ∩ R
ge
/n, R
ge
where I is the maximal interval of the existence of zr.
Proof. Consider αr−arctan z
r and remark that for r ∈ I ∩ R
ge
/n, R
ge
the following
relations hold:
α
r
1
cos α
r
·
p
e
γ
−
ρ · g ·z
r
γ
−
sin α
r
r
≥
1
cos α
r
·
ρ · g · R
ge
n − 1
γ ·n
· tan α
c
n
R
ge
−
ρ · g ·R
ge
n − 1
γ ·n
· tan α
c
−
n
R
ge
≥ 0.
2.35
Hence: z
r−1/cos
2
αr ·α
r ≤ 0forr ∈ I ∩R
ge
/n, R
ge
.
3. Meniscus Inner Free Surface Shape Analysis in
the Case of Tube Growth
Consider now the differential equation
z
ρ · g ·z −p
i
γ
1
z
2
3/2
−
1
r
·
1
z
2
· z
for r ∈
R
gi
,
R
gi
R
ge
2
, 3.1
and α
c
,α
g
such that 0 <α
c
<π/2 −α
g
, α
g
∈ 0,π/2.
12 Journal of Inequalities and Applications
Definition 3.1. A solution z zx of 3.1 describes the inner free surface of a static meniscus
on the interval R
gi
,r
i
R
gi
<r
i
< R
gi
R
ge
/2 if possesses the following properties:
a z
R
gi
tan α
c
,
b z
r
i
tanπ/2 −α
g
, and
c zR
gi
0andzr is strictly increasing on R
gi
,r
i
.
The described inner free surface is convex on R
gi
,r
i
if in addition the following inequality
holds:
d z
r > 0 ∀r ∈ R
gi
,r
i
.
Theorem 3.2. If there exists a solution of 3.1, which describes a convex inner free surface of a static
meniscus on the closed interval R
gi
,r
i
, then the following inequalities hold:
− γ ·
π/2 −
α
c
α
g
r
i
− R
gi
· cos α
c
−
γ
R
gi
· cos α
g
≤ p
i
≤−γ ·
π/2 −
α
c
α
g
r
i
− R
gi
· sin α
g
ρ · g ·
r
i
− R
gi
· tan
π
2
− α
g
−
γ
r
i
· sin α
c
.
3.2
Proof. Let zr be a solution of 3.1, which describes a convex inner free surface of a static
meniscus on the closed interval R
gi
,r
i
and αrarctan z
r. The function αr verifies the
equation
α
r
ρ · g · z
r
− p
i
γ
·
1
cos α
r
−
1
r
· tan α
r
3.3
and the boundary conditions
α
R
gi
α
c
,α
r
i
π
2
− α
g
. 3.4
Hence, there exists r
∈ R
gi
,r
i
such that the following equality holds:
p
i
−γ ·
π/2 −
α
c
α
g
r
i
− R
gi
· cos α
r
ρ · g · z
r
−
γ
r
· sin α
r
. 3.5
Since z
r > 0onR
gi
,r
i
, z
r is strictly increasing and αrarctan z
r is strictly
increasing on R
gi
,r
i
, therefore the following inequalities hold:
α
c
≤ α
r
≤
π
2
− α
g
,
sin α
g
≤ cos α
r
≤ cos α
c
,
sin α
c
≤ sin α
r
≤ cos α
g
,
ρ · g ·
r
− R
gi
· tan α
c
≤ ρ · g · z
r
≤ ρ · g ·
r
− R
gi
· tan
π
2
− α
g
.
3.6
Equality 3.5 and inequalities 3.6 imply inequalities 3.2.
Journal of Inequalities and Applications 13
Corollary 3.3. If r
i
m · R
gi
with 1 <m<R
gi
R
ge
/2 · R
gi
, then inequalities 3.2 become
− γ ·
π/2 −
α
c
α
g
m − 1
· R
gi
· cos α
c
−
γ
R
gi
· cos α
g
≤ p
i
≤−γ ·
π/2 −
α
c
α
g
m − 1
· R
gi
· sin α
g
ρ · g · R
gi
·
m − 1
· tan
π
2
− α
g
−
γ
m · R
gi
· sin α
c
.
3.7
Corollary 3.4. If m → R
gi
R
ge
/2 · R
ge
,thenr
i
→ R
gi
R
ge
/2 and 3.7 becomes
− 2 ·γ ·
π/2 −
α
c
α
g
R
ge
− R
gi
· cos α
c
−
γ
R
gi
· cos α
g
≤ p
i
≤−2 ·γ ·
π/2 −
α
c
α
g
R
ge
− R
gi
· sin α
g
ρ · g ·
R
ge
− R
gi
2
· tan
π
2
− α
g
−
2 · γ
R
gi
R
ge
· sin α
c
.
3.8
If m → 1,thenr
i
→ R
gi
and p
i
→−∞.
Theorem 3.5. Let m be such that 1 <m<R
gi
R
ge
/2 · R
gi
.Ifp
i
satisfies the inequality
p
i
< −γ ·
π/2 −
α
c
α
g
m − 1
· R
gi
· cos α
c
γ
R
gi
· cos α
g
, 3.9
then there exists r
i
in the closed interval R
gi
,m·R
gi
such that the solution of the initial value problem
z
ρ · g ·z −p
i
γ
·
1
z
2
3/2
−
1
r
·
1
z
2
· z
for R
gi
<r≤
R
gi
R
ge
2
,
z
R
gi
0,z
R
gi
tan α
c
, 0 <α
c
<
π
2
− α
g
,α
g
∈
0,
π
2
3.10
on the interval R
gi
,r
i
describes the convex inner free surface of a static meniscus.
Proof. Consider the solution zr of the initial value problem 3.10. Denote by I the maximal
interval on which the function zr exists and by αr the function αrarctan z
r defined
on I. Remark that for αr the following equality holds:
α
r
1
cos α
r
·
ρ · g ·z
r
− p
i
γ
−
sin α
r
r
. 3.11
14 Journal of Inequalities and Applications
Since
z
R
gi
> 0,z
R
gi
tan α
c
> 0,z
R
gi
< tan
π
2
− α
g
, 3.12
there exists r
∈ I, R
gi
<r
< R
gi
R
ge
/2 such that for any r ∈ r
,R
gi
the following
inequalities hold:
z
r
> 0,z
r
> 0,z
r
< tan
π
2
− α
g
. 3.13
Let r
∗
be defined by
r
∗
sup
r
∈I | R
gi
<r
<
R
gi
R
ge
2
such that for any r ∈
R
gi
,r
inequalities
3.13
hold
.
3.14
It is clear that r
∗
≤ R
gi
R
ge
/2 and for any r ∈ R
gi
,r
∗
inequalities 3.13 hold. Moreover,
z
r
∗
− 0lim
r →r
∗
,r<r
∗
z
r exists and satisfies, z
r
∗
− 0 > 0andz
r
∗
− 0 ≤ tanπ/2 − α
g
.
Hence zr
∗
−0lim
r →r
∗
,r<r
∗
zr is finite, it is strictly positive, and for every r ∈ R
gi
,r
∗
the
following inequalities hold:
tan α
c
≤ z
r
∗
− 0
≤ tan
π
2
− α
g
,
r
∗
− R
gi
· tan α
c
≤ z
r
∗
− 0
≤
r
∗
− R
gi
· tan
π
2
− α
g
.
3.15
We will show now that r
∗
≤ m · R
gi
and z
r
∗
− 0tanπ/2 −α
g
.
In order to show that r
∗
≤ m · R
gi
, we assume the contrary, that is, r
∗
>m· R
gi
. Under
this hypothesis we have
α
r
∗
− 0
− α
R
gi
α
r
·
r
∗
− R
gi
1
cos α
r
·
−
p
i
γ
ρ · g ·z
r
γ
−
sin α
r
r
·
r
∗
− R
gi
>
r
∗
− R
gi
cos α
r
·
π/2 −
α
c
α
g
R
gi
·
1
m − 1
· cos α
c
1
R
gi
· cos α
g
ρ · g ·z
r
γ
−
sin α
r
r
>
π
2
−
α
c
α
g
3.16
for some r
∈ R
gi
,r
∗
. Hence αr
∗
− 0 >π/2 − α
g
and it follows that there exists r
1
such
that R
gi
<r
1
<r
∗
and αr
1
π/2 − α
g
. This last inequality is impossible according to the
definition of r
∗
.
Journal of Inequalities and Applications 15
Therefore, r
∗
defined by 3.14 satisfies r
∗
<m·R
gi
.
Inordertoshowthatz
r
∗
− 0tanπ/2 − α
g
we remark that from the definition
3.14 of r
∗
it follows that at least one of the following three equalities holds:
z
r
∗
− 0
tan α
c
,z
r
∗
− 0
tan
π
2
− α
g
,z
r
∗
− 0
0. 3.17
Since z
r
∗
− 0 >z
r > tanα
c
for r ∈ R
gi
,r
∗
, it follows that the equality z
r
∗
− 0tan α
c
is impossible.
Hence we obtain that in r
∗
at least one of the following two equalities holds:
z
r
∗
− 0
0,z
r
∗
− 0
tan
π
2
− α
g
. 3.18
We show now that the equality z
r
∗
− 00 is impossible. For this we assume the contrary,
that is, z
r
∗
− 00. Under this hypothesis, from 3.11, we have
p
i
ρ · g · z
r
∗
− 0
−
γ
r
∗
· sin α
r
∗
− 0
> −
γ
R
gi
· cos α
g
> −
γ
R
gi
· cos α
g
− γ ·
1
m − 1
·
π/2 −
α
c
α
g
R
gi
· cos α
c
>p
i
,
3.19
what is impossible.
In this way we obtain that the equality z
r
∗
− 0tanπ/2 −α
g
holds.
For r
i
r
∗
the solution of the initial value problem 3.10 on the interval R
gi
,r
i
describes a convex inner free surface of a static meniscus.
Corollary 3.6. If for p
i
the following inequality holds
p
e
< −2 · γ ·
π/2 −
α
c
α
g
R
ge
− R
gi
· cos α
c
−
γ
R
gi
· cos α
g
, 3.20
then there exists r
i
in the interval R
gi
, R
gi
R
ge
/2 such that the solution of the initial value
problem 3.10 on the interval R
gi
,r
i
describes a convex inner free surface of a static meniscus.
Corollary 3.7. If for 1 <m
<m<R
gi
R
ge
/2 · R
gi
the following inequalities hold:
− γ ·
π/2 −
α
c
α
g
m
− 1
· R
gi
· sin α
g
ρ · g · R
gi
·
m
− 1
· tan
π
2
− α
g
−
γ
m
· R
gi
· sin α
c
<p
i
< −γ ·
π/2 −
α
c
α
g
m − 1
· R
gi
· cos α
c
−
γ
R
gi
· cos α
g
,
3.21
16 Journal of Inequalities and Applications
then there exists r
i
in the interval m
·R
gi
,m·R
gi
such that the solution of the initial value problem
3.10 on the interval R
gi
,r
i
describes a convex inner free surface of a static meniscus.
The existence of r
i
and the inequality r
i
≤ m · R
gi
follows from Theorem 3.5.The
inequality r
i
≥ m
· R
gi
follows from the Corollary 3.3.
Theorem 3.8. If a solution z
1
z
1
r of 3.1 describes a convex inner free surface of a static
meniscus on the interval R
gi
,r
i
r
i
∈ R
gi
, R
gi
R
ge
/2, then it is a weak minimum for the
energy functional of the melt column:
I
i
z
r
i
R
gi
γ ·
1
z
2
1/2
1
2
· ρ · g · z
2
− p · z
· r · dr,
z
R
gi
z
1
R
gi
0,z
r
i
z
1
r
i
.
3.22
Proof. It is similar to the proof of Theorem 2.9.
Definition 3.9. A solution z zr of 3.1 which describes the inner free surface of a static
meniscus is said to be stable if it is a weak minimum of the energy functional of the melt
column.
Remark 3.10. Theorem 3.8 shows that if z zr describes a convex inner free surface of a
static meniscus on the interval R
gi
,r
i
, then it is stable.
Remark 3.11. The solution of the initial value problem 3.10 is convex at R
gi
i.e., z
R
gi
> 0
if and only if
p
i
< −
γ
R
gi
· sin α
c
. 3.23
Theorem 3.12. If zr represents the i nner free surface of a static meniscus on the closed interval
R
gi
,r
i
r
i
< R
gi
R
ge
/2 which possesses the following properties:
a zr is convex at R
gi
, and
b the shape of zr changes once on the interval R
gi
,r
i
, that is, there exists a point r
∈
R
gi
,r
i
such that z
r > 0 for r ∈ R
gi
,r
i
,z
r
0andz
r < 0 for r ∈ r
,r
i
,
then there exists r
1
i
∈ R
gi
,r
i
such that z
r
1
i
tanπ/2 − α
g
and for p
i
the following inequality
holds:
−
γ
R
gi
<p
i
< −
γ
R
gi
· sin α
c
. 3.24
Proof. Since αrarctan z
r increases on R
gi
,r
and decreases on r
,r
i
,andαr
i
π/2 −α
g
>α
c
, there exists r
1
i
∈ R
gi
,r
i
such that αr
1
i
π/2 −α
g
. The maximum value αr
is less than π/2. From 3.11 we have
p
i
ρ · g · z
r
−
γ
r
· sin α
r
, 3.25
Journal of Inequalities and Applications 17
0
2
4
6
8
10
×10
−4
z m
0.075 0.0751 0.0752 0.0753 0.0754 0.0755
r m
p
e
−700 Pa
p
e
−1000 Pa
p
e
−1100 Pa
Figure 2: z versus r for p
e
−1100; −1000; −700Pa.
0.5
0.75
1
1.25
1.5
1.75
α rad
0.075 0.0751 0.0752 0.0753 0.0754 0.0755
r m
p
e
−700 Pa p
e
−1000 Pa
p
e
−1100 Pa
Figure 3: α versus r for p
e
−1100; −1000; −700Pa.
and therefore
−
γ
R
gi
<p
i
< −
γ
R
gi
· sin α
c
. 3.26
Remark 3.13. If the solution zr of the initial value problem 3.10 is concave i.e., z
r < 0
on the interval R
gi
,r
i
r
i
∈ R
gi
, R
gi
R
ge
/2, then it does not describe the inner free
surface of a static meniscus on R
gi
,r
i
.
18 Journal of Inequalities and Applications
0
2
4
6
8
10
×10
−4
z m
0.075 0.0751 0.0752 0.0753 0.0754 0.0755
r m
p
e
−800 Pa
p
e
−900 Pa
p
e
−1000 Pa
Figure 4: z versus r for p
e
−1000; −900; −800Pa.
Theorem 3.14. Let m be 1 <m<R
gi
R
ge
/2 · R
gi
.Ifforp
i
the following inequality holds:
p
i
>
γ
R
gi
ρ · g ·
m − 1
· R
gi
· tan
π
2
− α
g
, 3.27
then the solution zr of the initial value problem 3.10 is concave on the interval I ∩ R
gi
,m· R
gi
where I is the maximal interval of the existence of zr.
Proof. Consider αrarctan z
r and remark that for r ∈ I ∩ R
gi
,m· R
gi
the following
relations hold:
α
r
1
cos α
r
·
ρ · g · z
r
γ
−
p
i
γ
−
sin α
r
≤
1
cos α
r
·
ρ · g ·
m − 1
· R
gi
γ
· tan
π
2
− α
g
−
n
R
gi
−
ρ · g ·
m − 1
· R
gi
γ
· tan
π
2
− α
g
−
sin α
r
< 0.
3.28
Hence z
r−1/cos
2
αr ·α
r < 0forr ∈ I ∩R
gi
,m· R
ge
.
Journal of Inequalities and Applications 19
0.5
0.75
1
1.25
1.5
1.75
α rad
0.075 0.0751 0.0752 0.0753 0.0754 0.0755
r m
p
e
−800 Pa
p
e
−900 Pa
p
e
−1000 Pa
Figure 5: α versus r for p
e
−1000; −900; −800Pa.
0
2
4
6
8
10
×10
−4
z m
0.075 0.0751 0.0752 0.0753 0.0754 0.0755
r m
p
e
−1000 Pa
p
e
−2000 Pa
p
e
−2500 Pa
Figure 6: z versus r for p
e
−2500; −2000; −1000Pa.
4. Numerical Illustration
Numerical computations were performed for a Si cylindrical tube for the following data:
R
ge
75.5 · 10
−3
m
,R
gi1
74.5 ·10
−3
m
,α
c
0.523
rad
,α
g
0.192
rad
,
ρ 2.5 · 10
3
kg/m
3
,γ 7.2 · 10
−1
N/m
,g 9.81
m/s
2
,
n 1.006,n
1.0006,m 1.0067,m
1.0006.
4.1
20 Journal of Inequalities and Applications
0.5
0.75
1
1.25
1.5
1.75
α rad
0.075 0.0751 0.0752 0.0753 0.0754 0.0755
r m
p
e
−2500 Pa
p
e
−2000 Pa
p
e
−1000 Pa
Figure 7: α versus r for p
e
−2500; −2000; −1000Pa.
0
16
32
48
64
80
×10
−4
z m
0.0685 0.0699 0.0713 0.0727 0.0741 0.0755
r m
p
e
3 Pa
p
e
4 Pa
p
e
2 Pa
Figure 8: z versus r for p
e
2; 3; 4Pa.
The objective was to verify if the necessary conditions are also sufficient, or if the
sufficient conditions are also necessary. Moreover, the above data were used in experiments
and the computed results can be tested against the experiments in order to evaluate the
accuracy of the theoretical predictions. This test is not the subject of this paper.
For the considered numerical data, inequality 2.7 becomes
−1179.443
Pa
≤ p
e
≤−194.682
Pa
. 4.2
Integration of 2.10 shows that for p
e
−1100; −1000 Pa there exists r
e
∈ R
ge
/n, R
ge
such that the solution is a convex outer free surface of a static meniscus on r
e
,R
ge
,butfor
Journal of Inequalities and Applications 21
0.5
0.75
1
1.25
1.5
1.75
α rad
0.0685 0.0699 0.0713 0.0727 0.0741 0.0755
r m
p
e
3 Pa
p
e
4 Pa
p
e
2 Pa
Figure 9: α versus r for p
e
2; 3; 4Pa.
0
2
4
6
8
10
×10
−3
z m
0.07 0.0711 0.0722 0.0733 0.0744 0.0755
r m
p
e
4.768 Pa
p
e
5 Pa
p
e
10 Pa
p
e
100 Pa
Figure 10: z versus r for p
e
4.768; 5; 10; 100 Pa.
p
e
−700Pa there is no r
e
∈ R
ge
/n, R
ge
such that the solution is a convex outer free
surface of a static meniscus on r
e
,R
ge
Figures 2 and 3. Hence, inequality 2.7 is not a
sufficient condition.
For the same numerical data inequality 2.9 becomes p
e
< −1179.443 Pa. We have
already obtained that for p
e
−1100; −1000 Pa there exists r
e
in the interval R
ge
/n, R
ge
such that the solution of 2.10 describes a convex outer free surface of a static meniscus on
the interval r
e
,R
ge
. Hence, inequality 2.9 is not a necessary condition.
For the same numerical data inequality 2.19 becomes p
e
< −1061.728 Pa.
Integration shows that for p
e
−1000; −900; −800 Pa there exists r
e
∈ R
gi
R
ge
/2,R
ge
,
such that the solution of 2.10 describes a convex outer free surface of a static meniscus
22 Journal of Inequalities and Applications
0.5
0.75
1
1.25
1.5
1.75
α rad
0.07 0.0711 0.0722 0.0733 0.0744 0.0755
r m
p
e
4.768 Pa
p
e
5 Pa
p
e
10 Pa
p
e
100 Pa
Figure 11: α versus r for p
e
4.768; 5; 10; 100 Pa.
0
2
4
6
8
10
×10
−3
z m
0.0685 0.0699 0.0713 0.0727 0.0741 0.0755
r m
p
e
15 Pa
Figure 12: z versus r for p
e
15 Pa.
on the closed interval r
e
,R
ge
Figures 4 and 5. Hence, inequality 2.19 is not a necessary
condition.
For the same numerical data inequality 2.20 becomes −2580 <p
e
< −1179.443Pa.
Integration of 2.10 illustrates the above phenomenon for p
e
−2000; −2500 PaFigures
6 and 7 and also the fact that the condition is not necessary see p
e
−1000 Pa.
For the same numerical data inequality 2.21 becomes p
e
< 4.768 Pa. Integration of
2.10 illustrates the above phenomenon for p
e
2; 3; 4 PaFigures 8 and 9.
For the considered numerical data inequality 2.32 becomes 4.768 Pa <p
e
<
66.23 Pa. Integration shows that for p
e
4.768 Pa there is no r
e
∈ R
ge
/n, R
ge
such that the
solution of 2.10 is a nonglobally convex outer free surface of a static meniscus on r
e
,R
ge
.
Journal of Inequalities and Applications 23
0.5
0.75
1
1.25
1.5
1.75
α rad
0.0685 0.0699 0.0713 0.0727 0.0741 0.0755
r m
p
e
15 Pa
Figure 13: α versus r for p
e
15 Pa.
0
2
4
6
8
10
×10
−4
z m
0.0745 0.0746 0.0747 0.0748 0.0749 0.075
r m
p
i
−900 Pa
p
i
−1000 Pa
p
i
−550 Pa
Figure 14: z versus r for p
i
−1000; −900; −550Pa.
Moreover, for p
e
5; 10; 100 Pa it is not anymore the outer free surface of a static meniscus
Figures 10 and 11. Hence, inequality 2.32 is not a sufficient condition.
For the same numerical data inequality 2.33 becomes 4.768Pa ≤ p
e
≤ 72.5 Pa.
We have already obtained that for p
e
5; 10 PaFigure 10 the solution of 2.10 is not
anymore the outer free surface of a static meniscus. Hence, inequality 2.33 is not a sufficient
condition.
For the same numerical data inequality 2.34 becomes p
e
> 15.9696 Pa. Integration
of 2.10 for p
e
15 Pa proves that the solution is globally concave on R
ge
/n, R
ge
Figures
12 and 13. Hence, inequality 2.34 is not a necessary condition.
24 Journal of Inequalities and Applications
0.5
0.75
1
1.25
1.5
1.75
α rad
0.0745 0.0746 0.0747 0.0748 0.0749 0.075
r m
p
i
−900 Pap
i
−1000 Pa
p
i
−550 Pa
Figure 15: α versus r for p
i
−1000; −900; −550Pa.
0
2
4
6
8
10
×10
−4
z m
0.0745 0.0746 0.0747 0.0748 0.0749 0.075
r m
p
i
−900 Pa
p
i
−1000 Pa
p
i
−800 Pa
Figure 16: z versus r for p
i
−1000; −900; −800Pa.
For the considered numerical data inequality 3.7 becomes −1076.682 Pa ≤ p
i
≤
−177.201 Pa. Integration of 3.10 shows that for p
i
−1000; −900 Pa there exists r
i
∈
R
gi
,m · R
gi
such that the solution is a convex inner free surface of a static meniscus on
R
gi
,r
i
,butforp
i
−550 Pa there is no r
i
∈ R
gi
,m· R
gi
such that the solution is a convex
inner free surface of a static meniscus on R
gi
,r
i
Figures 14 and 15. Hence, inequality 3.7
is not a sufficient condition.
For the considered numerical data 3.9 becomes p
i
< −1076.682 Pa. We have already
obtained that for p
i
−1000; −900 Pa there exists r
i
∈ R
gi
,m· R
gi
such that the solution of
3.10 describes a convex inner free surface of a static meniscus on R
gi
,r
i
. Hence, inequality
3.9 is not a necessary condition.
Journal of Inequalities and Applications 25
0.5
0.75
1
1.25
1.5
1.75
α rad
0.0745 0.0746 0.0747 0.0748 0.0749 0.075
r m
p
i
−900 Pa
p
i
−1000 Pa
p
i
−800 Pa
Figure 17: α versus r for p
i
−1000; −900; −800Pa.
0
2
4
6
8
10
×10
−4
z m
0.0745 0.0746 0.0747 0.0748 0.0749 0.075
r m
p
i
−1500 Pa
p
i
−2000 Pa
p
i
−2500 Pa
p
i
−500 Pa
Figure 18: z versus r for p
i
−2500; −2000; −1500; −500 Pa.
For the considered numerical data 3.20 becomes p
i
< −1075.98 Pa. Integration of
3.10 shows that for p
i
−1000; −900; −800 Pa there exists r
i
∈ R
gi
, R
gi
R
ge
/2 such
that the solution describes a convex inner free surface of a static meniscus on R
gi
,r
i
Figures
16 and 17. Hence, inequality 3.20 is not a necessary condition.
For the considered numerical data 3.21 becomes −2613 Pa <p
i
< −1076.68Pa.
Integration of 3.10 illustrates the above phenomenon for p
i
−1500; −2000; −2500 Pa and
also the fact that the condition is not necessary Figures 18 and 19see p
i
−500 Pa.
For the same numerical data inequality 3.23 becomes p
i
< −4.8322 Pa. Integration
of 3.10 illustrates the above phenomenon for p
i
−7; −6; −5 PaFigures 20 and 21.
For the considered numerical data inequality 3.24 becomes −9.664 Pa <p
i
<
−4.827 Pa. Integration shows that for p
i
−5 Pa there exists r
i
∈ R
gi
,m · R
gi
such
